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FINAL EXAMACSC12/71-200MATHEMATICAL STATISTICS

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FINAL EXAM

ACSC12/71-200

MATHEMATICAL STATISTICS

Question 1: (8 Total Marks)

Indicate whether each statement is TRUE or FALSE. [2 marks each] a) If two independent random variables, π and π, have distributions with cgfs ππ(π‘) and ππ(π‘), respectively, then the distribution of π = π + π has cgf ππ(π‘) = ππ(π‘) + ππ(π‘).

1. If π and π are independent normally distributed random variables, then their sum, π = π + π, and their difference, π· = π − π, are also independent.
2. If π is a discrete random variable with support {0,1,2 … } and πΈ(π) = π, a positive integer, and π1, π2, … are independent and identically distributed random variables all independent of π, then the random sum π ππ and the fixed sum π = ∑ππ=1 ππ must have πππ(π) ≥ πππ(π).
3. If π1 and π2 are estimators of a parameter π based on the same data, and π1 is unbiased but π2 is

not, then the variance of π1 must be lower than the variance of π2.

Question 2: (10 Total Marks)

Let π1, … , π100 be iid πππππ’ππ(2,1/π) random variables with πΈ π and pdf: 2     −π₯2

ππ

π

Suppose their observed values satisfy .

1. a) Find the method of moments estimator, πΜπππ, and the maximum likelihood estimator, πΜππΏπΈ, for

π based on this data. [NOTE: Recall  [4 marks] b) Find an approximate 95% confidence interval for π based on πΜππΏπΈ. [3 marks]

1. c) Find an approximate 95% confidence interval for π based on πΜπππ. [3 marks]

Question 3: (9 Total Marks)

The route I walk from my Gold Coast apartment to Bond has 4 street crossings. I have determined I independently encounter road traffic at a crossing with probability 0.4. Thus, the number of crossings I must stop and wait at during a trip is π ~ π΅ππ(4,0.4). Also, the lengths of time I have to stop are iid random variables, ππ, so the extra time added to my trip due to traffic is π = ∑ππ=1 ππ. Finally, I have discovered my walking time without any stops, π, is normally distributed with mean 8 minutes and standard deviation 2 minutes.

1. a) If the ππ’s are independent of π and πΈ(ππ) = 2 min, ππ·(ππ) = 0.5 min and πΆπππ(π, π) = 0.5, find the mean and standard deviation of my total trip time, π + π. [4 marks] b) Assume ππ has mgf ππ(π‘) and consider new random variables ππ defined by ππ= 0 if I don’t have to stop at the ith crossing and ππ= ππ if I do. Show π′ = ∑4π=1 ππ has the same distribution as π and comment. [HINT: Find mgfs of π and π′ and compare.] [2 marks] c) Now, suppose ππ ~ π(2,0.5) and πΆπππ(π1, π) = 0.8. Calculate πΈ(π|π1 = 3). [3 marks]

Question 4: (9 Total Marks)

You have recently been appointed Deputy Defective DooDad Detective at WhatsItWorks Corp and you have determined the proportion of defective DooDads (called DooDuds) from the current DooDad Delivery Device is 10%. Management is considering upgrading to the DoctorDoo2000, advertised as making fewer defectives, but say it is only worth it if you confirm the lower defective rate (at πΌ = 0.05).

1. You have a source on the DoctorDoo2000 development team who tells you its true defective rate is π = 9%. Based on this, what size sample will have a power of detection of 90% (i.e., how big does π need to be so that the chance of rejecting π»0: π = 0.1 is 90%)? [2 marks]
2. You take a sample of 7500 DooDads from a DoctorDoo2000 and observe 691 DooDuds, should you recommend the upgrade to management or not? [4 marks]
3. Your source now informs you DoctorDoo2000s may be inconsistent, individual machines having varying defective rates. This concern is based on samples from two DoctorDoo2000s, each of size 5000, the first had 455 DooDuds and the second 595. Do you agree this is statistically significant evidence (at πΌ = 0.05) of a difference in defective rates between machines? [3 marks]

Question 5: (9 Total Marks)

I am a huge fan of superhero comic books and movies; however, I have always liked Marvel characters better than DC characters. To assess whether my views are consistent with the wider superhero-loving public, I gathered data about the box-office opening weekend earnings (in millions of US dollars) for a number of recent superhero movies from both companies and summarised the data:

 Statistic DC Marvel Average, πΜ 111.96 135.25 Standard Deviation, π 40.66 73.36 π 8 23
1. a) Using this information, calculate a CLT-based 95% confidence interval for the mean difference in first weekend earnings for Marvel versus DC movies. [4 marks] b) Earnings are generally not symmetrically distributed, so there is concern about using CLT-based intervals with such small samples. To deal with this, a bootstrap-t procedure is implemented. From π΅ = 10,000 bootstrap samples, we calculate values of the form:

π‘π∗ =  ππππ£,π) − (πΜπ·πΆ − πΜππππ£)

π

and find that π‘ and π‘.  Use this information to construct a bootstrap-t confidence interval for the mean difference in weekend earnings. [2 marks]

Suppose overall earnings of superhero movies are independent and πΏπ(π, 1), so that, ππ, the earnings of a random superhero movie has expectation πΈ(ππ) = ππ+0.5 = π(π).  Define estimators π1 = πΜ = πln(πΜ) and π2 = πΜlnΜΜΜ(ΜπΜΜΜ) (i.e., “π to the average log value” instead of “π to the log of the average value”).

1. c) Clearly, π1 is an unbiased estimate of π(π). Show π2 is a biased and inconsistent estimator of π(π), but is still more accurate than π1 when the sample size is less than 10. [3 marks].

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